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Basic Atomic Question

  1. Apr 19, 2012 #1
    1. The problem statement, all variables and given/known data
    In question 50 in this link:

    http://www.tarleton.edu/Faculty/alow/prob21.htm

    I was wondering how the numbers 4.00260 amu + 1.00867 amu – 2.01410 amu – 3.01605 amu been obtained?

    Thank you.


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 19, 2012 #2

    tiny-tim

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  4. Apr 20, 2012 #3
    Thank you.

    The first part of question 50 states 2H + 3H → 4He + 1n

    So are they multiplying these numbers by a certain constant to arrive at the numbers of 4.00260 amu + 1.00867 amu – 2.01410 amu – 3.01605 amu

    If so, what is this constant?

    As the numbers do look just slightly more than the numbers which I have emboldened.
     
  5. Apr 20, 2012 #4

    tiny-tim

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    they're not a constant times anything, atomic weight is not a linear function of the number of nucleons

    fuse two protons (1H) together, that's two lots of 1.00867 = 2.01734, you get a 2H (= 2.01410) and some energy left over, which is the energy of fusion that they're trying to make use of in fusion reactors
     
  6. Apr 20, 2012 #5

    Borek

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    You may want to change it to something closer to reality :wink:
     
  7. Apr 20, 2012 #6

    tiny-tim

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